Question

Last Question of the day

The function f is continuous on the interval [4, 15], with some of its values given in the table above. Estimate the average value of the function with a Trapezoidal Approximation, using the 4 intervals between those given points.

x 4 9 11 14 15
f(x) –6 –11 –18 –21 –25

Answer 1

okay so i'v done this a couple of times but i keep getting the wrong answer.

Answer 2

you have a bunch of trapezoids, that is all.

Area of each trapezoid is:
\(\large\displaystyle A=(\Delta x)(h_1+h_2)/2\)

where \(\Delta x\) is the width
and \(\large\displaystyle (h_1+h_2)/2\) is the average height

\(\large\displaystyle Trapezoid_1=(9-4)(-6+-11)/2\)
\(\large\displaystyle Trapezoid_2=(11-9)(-18+-11)/2\)
\(\large\displaystyle Trapezoid_3=(14-11)(-21+-18)/2\)
\(\large\displaystyle Trapezoid_4=(15-14)(-25+-21)/2\)

Answer 3

then add all trapezoids.

Answer 4

Dont wonder the area is negative, that is because the entire thing is under the x-axis.

Answer 5

yep im getting the same answer i was getting before , still wrong

Answer 6

these are my answer choices
–12.727
–11.546
–16.273
–13.909

Answer 7

@SolomonZelman you still there ?

Answer 8

yes, i am here. I am just glitching outside the US a little.

Answer 9

what did you get when you added the trapezoids?

Answer 10

153, i did the same thing but by taking the left point and right point sum and averaging them

Answer 11

Yes, the area of the trapezoid is given by the product of its average height times the width, and the area of your full shape is the sum of all trapezoids...

i dont understand why these options are here.

Answer 12

i get -153 doing a completely valid method.... maybe your table has typos?

Answer 13

perhaps we're finding the wrong thing . it says find the average value might mean the average slope not average area under the curve

Answer 14

@ganeshie8 any idea?

Answer 15

\[\frac{1}{b-a} \int\limits _a^b f(x) dx\]

Answer 16

The average function value formula is: